When to use quadratic equations?

[struggtofunc]

Hello everyone! Apologies if this is a very repetitive question but I have gone through previous forum posts and am still struggling to understand how to identify which equations are appropriate. In the problem below, I have used the kinematic equation of "v = v(i) + at" but my answer is incorrect. Rather, I was told to use a quadratic equation. My question is, what determines whether to use a quadratic equation of not? I was under the impression that because I had the values of v(i), v, and a, that i could simply use the kinematic equation. Why would this be incorrect?

Again, I'm very sorry if I have posted this incorrectly. Any help would be much appreciated!

Thank you very much!

J

1. Homework Statement
A swimmer bounces straight up from a diving board and falls feet first into a pool. She starts with a velocity of 4.00 m/s, and her takeoff point is 1.80 m above the pool. (a) How long are her feet in the air?

y(i)=1.80m
y = 0m
a = -9.8 ms-2
v(i) = 4.00 ms-1

Homework Equations

y-y(i) = v(i)t + (1/2)(a)(t^2)
or
v = v(i) + at

The Attempt at a Solution

I have used the equation v = v(i) + at and calculated the answer as:

0 = 4.00 + (-9.8)(t)

t = (-4.00)/(-9.8)
t = 0.408 seconds

 

[tnich]

Hello everyone! Apologies if this is a very repetitive question but I have gone through previous forum posts and am still struggling to understand how to identify which equations are appropriate. In the problem below, I have used the kinematic equation of "v = v(i) + at" but my answer is incorrect. Rather, I was told to use a quadratic equation. My question is, what determines whether to use a quadratic equation of not? I was under the impression that because I had the values of v(i), v, and a, that i could simply use the kinematic equation. Why would this be incorrect?

Again, I'm very sorry if I have posted this incorrectly. Any help would be much appreciated!

Thank you very much!

J

1. Homework Statement
A swimmer bounces straight up from a diving board and falls feet first into a pool. She starts with a velocity of 4.00 m/s, and her takeoff point is 1.80 m above the pool. (a) How long are her feet in the air?

y(i)=1.80m
y = 0m
a = -9.8 ms-2
v(i) = 4.00 ms-1

Homework Equations

y-y(i) = v(i)t + (1/2)(a)(t^2)
or
v = v(i) + at

The Attempt at a Solution

I have used the equation v = v(i) + at and calculated the answer as:

0 = 4.00 + (-9.8)(t)

t = (-4.00)/(-9.8)
t = 0.408 seconds

You have calculated the time for her to reach zero vertical speed. At what speed does she enter the pool?

 

[struggtofunc]

Thanks for your reply! I think I see the problem now. So because the question is asking for time in the air, my V(f) value should be calculated just before she reaches rest. Thankyou!

 

[Marc Rindermann]

You have calculated the time for her to reach zero vertical speed. At what speed does she enter the pool?

In other words, you have calculated the time after which she is at the highest point. Now you have to add the time it takes to fall down into the pool.

Thanks for your reply! I think I see the problem now. So because the question is asking for time in the air, my V(f) value should be calculated just before she reaches rest. Thankyou!

she won't be at rest when she hits the water. she'll be at her highest speed just before she hits the water.

 

[struggtofunc]

Thankyou for your help! Would it be appropriate to solve this in two steps? One to reach the peak of trajectory motion and one to reach the pool?

 

[tnich]

Thankyou for your help! Would it be appropriate to solve this in two steps? One to reach the peak of trajectory motion and one to reach the pool?

Consider how you could calculate her velocity when her feet touch the water.

 

[Charles Link]

To the OP: You know the position $y$ where her feet are as they touch the water, i.e. basically you know $y_{water}-y_{initial}$. What equation might work best, given that information?

 

[tnich]

Consider how you could calculate her velocity when her feet touch the water.

I think it would be much easier to use the other equation.

 

[struggtofunc]

Thanks for all the help everyone!

I've calculated the time using two equations rather than quadratics:

v^2 = v(i)^2 + 2a(x-x(i))
v^2 = (4.00)^2 + 2(-9.8)(-1.80)

v = square root of 51.28 = +- 7.16 ms-1

v = v(i) + at
t = (+-7.16 - 4.00)/(-9.8)
t = - 0.32 s
t = 1.14 s

Thank you all so much! J

 

[Marc Rindermann]

Thankyou for your help! Would it be appropriate to solve this in two steps? One to reach the peak of trajectory motion and one to reach the pool?

it could certainly be done in more than one step. However, now that you have the time it takes to get up to the highest point of her jump you'd need to calculate how high she's up in the air. you need the information to calculate the time it takes to fall from this height, given that the speed at this point is 0 m/s.

you could use the exact same equation to calculate it all in one step. with initial height that of the board and initial velocity when she jumps off the board.

 

[Charles Link]

Suggestion is to use the distance equation that is quadratic in time, especially if you haven't had much practice with quadratic equations. It's a good way to get practice with solving a quadratic equation. I didn't check your answer yet using the method that you used, but you should get the same answer using the distance equation $y-y_i=v_i t +(\frac{1}{2})at^2$. $\\$ Editing: I have now checked your solution, and it is correct. And I also got the same answer using the above distance equation.

 

[tnich]

Thanks for all the help everyone!

I've calculated the time using two equations rather than quadratics:

v^2 = v(i)^2 + 2a(x-x(i))
v^2 = (4.00)^2 + 2(-9.8)(-1.80)

v = square root of 51.28 = +- 7.16 ms-1

v = v(i) + at
t = (+-7.16 - 4.00)/(-9.8)
t = - 0.32 s
t = 1.14 s

Thank you all so much! J

I can't see any fault with that. You have done exactly the same calculations that would have been required to solve the quadratic.